Question: $g(x) = 6x^{3}+4x-3-4(h(x))$ $f(n) = 2n^{2}-n$ $h(x) = 5x+7+3(f(x))$ $ f(h(-1)) = {?} $
First, let's solve for the value of the inner function, $h(-1)$ . Then we'll know what to plug into the outer function. $h(-1) = (5)(-1)+7+3(f(-1))$ To solve for the value of $h$ , we need to solve for the value of $f(-1)$ $f(-1) = 2(-1)^{2}-(-1)$ $f(-1) = 3$ That means $h(-1) = (5)(-1)+7+(3)(3)$ $h(-1) = 11$ Now we know that $h(-1) = 11$ . Let's solve for $f(h(-1))$ , which is $f(11)$ $f(11) = 2(11^{2})-11$ $f(11) = 231$